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Linear Algebra : Linear Algebra

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Example Questions

Example Question #42 : Matrices

Find the product $A \times B$ .

$A=\begin{bmatrix} -1 &5&2\\ 4 &-4&5 \end{bmatrix}$ , $B=\begin{bmatrix} 2 &-1&5\\ 0 &-7&4 \end{bmatrix}$

Possible Answers:

$A \times B=\begin{bmatrix} -3 &6&-3\\ 4 &3&1 \end{bmatrix}$

$A \times B=\begin{bmatrix} -2 &-5&10\\ 0 &28&20 \end{bmatrix}$

$A \times B$ cannot be multiplied.

$A \times B=\begin{bmatrix} 1 &4&7\\ 0 &-7&9 \end{bmatrix}$

Correct answer:

$A \times B$ cannot be multiplied.

Explanation:

If is an $(m\times n)$ matrix and is an $(r\times s)$ matrix,

$(A\times B)$ can only be multiplied if n=r , or the number of columns in equals the number of rows in . Otherwise there is a mismatch, and the two matrices can not be multiplied. $(A\times B)$ will be an $(m\times s)$ matrix

Since is a $2\times3$ matrix and is a $2\times3$ matrix, $n\neq r$ so $(A\times B)$ cannot be multiplied.

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Example Question #731 : Linear Algebra

$A = \begin{bmatrix} \frac{1}{4} & \frac{\sqrt{3}}{4} \\ \\ \frac{\sqrt{3}}{4} & \frac{3}{4} \end{bmatrix}$

True or false:

is an example of an idempotent matrix.

Possible Answers:

False

True

Correct answer:

True

Explanation:

is an idempotent matrix, by definition, if $A^{2} = A$ . Multiply by itself by multiplying rows by columns - multiplying elements in corresponding positions and adding the products:

$A ^{2}= \begin{bmatrix} \frac{1}{4} & \frac{\sqrt{3}}{4} \\ \\ \frac{\sqrt{3}}{4} & \frac{3}{4} \end{bmatrix} \begin{bmatrix} \frac{1}{4} & \frac{\sqrt{3}}{4} \\ \\ \frac{\sqrt{3}}{4} & \frac{3}{4} \end{bmatrix}$

$= \begin{bmatrix} \frac{1}{4} \cdot \frac{1}{4} + \frac{\sqrt{3}}{4} \cdot \frac{\sqrt{3}}{4} & \frac{1}{4} \cdot \frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} \cdot \frac{3}{4} \\ \\\ \ \frac{\sqrt{3}}{4} \cdot \frac{1}{4} + \frac{3}{4} \cdot \frac{\sqrt{3}}{4} & \frac{\sqrt{3}}{4} \cdot \frac{\sqrt{3}}{4} + \frac{3}{4} \cdot \frac{3}{4} \end{bmatrix}$

$= \begin{bmatrix} \frac{1}{16} + \frac{ 3}{16} & \frac{\sqrt{3}}{16} + \frac{3\sqrt{3}}{16} \\ \\\ \ \frac{\sqrt{3}}{16} +\frac{3\sqrt{3}}{16} & \frac{3}{16} + \frac{9}{16} \end{bmatrix}$

$= \begin{bmatrix} \frac{1}{4} & \frac{\sqrt{3}}{4} \\ \\ \frac{\sqrt{3}}{4} & \frac{3}{4} \end{bmatrix} =A$ .

$A^{2} = A$ , making idempotent.

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Example Question #732 : Linear Algebra

$A = \begin{bmatrix} 0 &- 1 \\ 1 & 0 \end{bmatrix}$

True or false:

is an example of an idempotent matrix.

Possible Answers:

False

True

Correct answer:

False

Explanation:

is an idempotent matrix, by definition, if $A^{2} = A$ . Multiply by itself by multiplying rows by columns - multiplying elements in corresponding positions and adding the products:

$A^{2} = \begin{bmatrix} 0 &- 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 &- 1 \\ 1 & 0 \end{bmatrix}$

$= \begin{bmatrix} 0(0) + (-1) (1) & 0 (-1)+ (-1) (0) \\ 1(0) + 0 (1) & 1(-1)+ 0(0) \end{bmatrix}$

$= \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}$

$A^{2} \ne A$ , so is not idempotent.

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Example Question #733 : Linear Algebra

For any given value , how many $n \times n$ nonsingular idempotent matrices exist?

Possible Answers:

Zero

Infinitely many

One

Two

Correct answer:

One

Explanation:

is nonsingular, by definition, if it has an inverse - that is, if $A^{-1}$ exists. is an idempotent matrix, by definition, if

$A^{2} = A$

Premultiplying both sides of the equation by $A^{-1}$ , we get

$A^{-1}(A^{2}) = A^{-1}A$

$A^{-1}(AA )= I$ ,

where is the $n \times n$ identity matrix.

Matrix multiplication is associative, so

$(A^{-1}A)A = I$

IA = I

A = I .

Therefore, the only nonsingular idempotent matrix of a given dimension $n \times n$ is the identity.

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Example Question #734 : Linear Algebra

Markov 2

The above diagram shows a board for a game of chance. A player moves according to the flip of a fair coin, depending on his current location. For example, if he is on the green square, he will move to the orange square if the coin comes up heads, and to the pink square if it comes up tails.

Construct a stochastic matrix that models this game, with the rows/columns representing, in order, the orange, pink, blue, and green squares.

Possible Answers:

$\begin{bmatrix} 0.5 & 0.5 & 0&0 \\ 0.5 &0 & 0.5 & 0\\ 0 & 0.5 &0 & 0.5 \\ 0.5 & 0 .5 & 0 & 0 \end{bmatrix}$

$\begin{bmatrix} 0 & 0.5 & 0&0.5 \\ 0.5 &0 & 0.5& 0 \\ 0 & 0.5 &0 & 0.5 \\ 0.5 & 0 & 0.5 & 0 \end{bmatrix}$

$\begin{bmatrix} 0.5 & 0.5 & 0&0.5 \\ 0.5 &0 & 0.5 & 0.5\\ 0 & 0.5 &0 & 0 \\ 0 & 0 & 0.5 & 0 \end{bmatrix}$

$\begin{bmatrix} 0.5 & 0 & 0.5&0.5 \\ 0.5 &0.5 & 0 & 0.5\\ 0 & 0 &0.5 & 0 \\ 0 & 0.5 & 0 & 0 \end{bmatrix}$

$\begin{bmatrix} 0.5 &0 & 0.5& 0 \\ 0 & 0.5 &0 & 0.5 \\ 0.5 & 0 & 0.5 & 0 \\ 0 & 0.5 & 0&0.5 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} 0.5 & 0.5 & 0&0.5 \\ 0.5 &0 & 0.5 & 0.5\\ 0 & 0.5 &0 & 0 \\ 0 & 0 & 0.5 & 0 \end{bmatrix}$

Explanation:

Since the coin is fair, each outcome will come up with probability 0.5. Therefore, a player on orange end up on orange or pink with 0.5 probability each; a player on pink will end up on orange or blue with probability 0.5 each; a player on blue will end up on pink or green with probability 0.5 each; and a player on green will end up on orange or pink with probability 0.5 each. The matrix that models this is

$\begin{bmatrix} 0.5 & 0.5 & 0&0.5 \\ 0.5 &0 & 0.5 & 0.5\\ 0 & 0.5 &0 & 0 \\ 0 & 0 & 0.5 & 0 \end{bmatrix}$ .

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Example Question #735 : Linear Algebra

Markov 2

It is recommended that you use a calculator with matrix arithmetic capability for this problem.

The player agrees to start on the pink square. Which square is he most likely to end up on after eight moves?

Possible Answers:

Orange

Pink

Green

Blue

Correct answer:

Orange

Explanation:

$P = \begin{bmatrix} 0.5 & 0.5 & 0&0.5 \\ 0.5 &0 & 0.5 & 0.5\\ 0 & 0.5 &0 & 0 \\ 0 & 0 & 0.5 & 0 \end{bmatrix}$ .

The matrix that models the probabilities that the player will end up on a given square, given that he starts on a particular square, is $P^{8}$ , which through calculation is

$P^{8} = \begin{bmatrix} 0.4180 & 0.4141&0.4180 &0.4180 \\ 0.3320 &0.3359 &0.3320 &0.3320 \\ 0.1680 & 0.1641 & 0.1680 &0.1680 \\ 0.0820 & 0.0859 &0.0820 & 0.0820 \end{bmatrix}$

Since the player is starting on pink, examine the second column; the greatest entry is in the second row, which represents ending up on orange.

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Example Question #51 : Matrix Matrix Product

True or false: A matrix whose determinant is neither 0 nor 1 cannot be an idempotent matrix.

Possible Answers:

False

True

Correct answer:

True

Explanation:

is an idempotent matrix, by definition, if $A^{2} = A$ . Since the determinant of the product of two matrices is equal to the product of their determinants, it follows that

$\det A^{2 }= (\det A )^{2}$

and, since $A^{2} = A$ ,

$\det A^{2 }= \det A$ .

By transitivity,

$(\det A )^{2} = \det A$ .

The only two numbers equal to their own squares are 0 and 1, so

$\det A = 0$ or $\det A = 1$ .

This makes the statement true.

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Example Question #732 : Linear Algebra

$A = \begin{bmatrix} 2+ i & 3 - i & i \\ -4 & 1+ i & 3 \end{bmatrix}$ .

Calculate $A ^{*}A$ .

Possible Answers:

$\begin{bmatrix}16 & -6-5i \\ -6-5i &25\end{bmatrix}$

$\begin{bmatrix} 21&1-9i & -11+2i \\1+9i & 12& 2\\ -11-2i &2& 10\end{bmatrix}$

$\begin{bmatrix}-9+4i &3-3i& -13+2i \\ 3-3i & 8-4i & 4+6i \\ -13+2i &4+6i& 8\end{bmatrix}$

$\begin{bmatrix}10-2i & -4+i \\ -4+i & 25+2i \end{bmatrix}$

$A ^{*}A$ is undefined.

Correct answer:

$\begin{bmatrix} 21&1-9i & -11+2i \\1+9i & 12& 2\\ -11-2i &2& 10\end{bmatrix}$

Explanation:

$A^{*}$ , the conjugate transpose of , can be found by first taking the transpose of :

$A = \begin{bmatrix} 2+ i & 3 - i & i \\ -4 & 1+ i & 3 \end{bmatrix}$ ,

$A^{T} = \begin{bmatrix} 2+ i & -4 \\ 3- i & 1+ i \\i & 3 \end{bmatrix}$

then changing each element to its complex conjugate:

$A^{*} = \begin{bmatrix} 2- i & -4 \\ 3+i & 1- i \\-i & 3 \end{bmatrix}$

Find the product $A ^{*}A$ by multiplying the rows of $A^{*}$ by the columns of ; that is, add the product of the terms in corresponding positions:

$A ^{*}A$

$= \begin{bmatrix} 2- i & -4 \\ 3+ i & 1- i \\-i & 3 \end{bmatrix} \begin{bmatrix} 2+ i & 3 - i & i \\ -4 & 1+ i & 3 \end{bmatrix}$

$= \begin{bmatrix} (2- i) (2+ i) + (-4)(-4) &(2- i) (3 - i) + (-4)(1+i) & (2- i) ( i) + (-4)(3) \\ (3+i) (2+ i) + (1-i)(-4) & (3+i) (3 - i) +(1-i)(1+i) & (3+i) ( i) +(1-i)(3) \\ ( -i) (2+ i) + (3)(-4) &(- i) (3 - i) + (3)(1+i) & (- i) ( i) + (3)(3)\end{bmatrix}$

$= \begin{bmatrix} 21&1-9i & -11+2i \\1+9i & 12& 2\\ -11-2i &2& 10\end{bmatrix}$

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Example Question #732 : Linear Algebra

$A = \begin{bmatrix} 2+ i & 3 - i & i \\ -4 & 1+ i & 3 \end{bmatrix}$

Calculate $AA^{T}$ .

Possible Answers:

$\begin{bmatrix}-9+4i &3-3i& -13+2i \\ 3-3i & 8-4i & 4+6i \\ -13+2i &4+6i& 8\end{bmatrix}$

$\begin{bmatrix}16 & -6-5i \\ -6-5i &25\end{bmatrix}$

$\begin{bmatrix}10-2i & -4+i \\ -4+i & 25+2i \end{bmatrix}$

$AA^{T}$ is undefined.

$\begin{bmatrix} 21&1-9i & -11+2i \\1+9i & 12& 2\\ -11-2i &2& 10\end{bmatrix}$

Correct answer:

$\begin{bmatrix}10-2i & -4+i \\ -4+i & 25+2i \end{bmatrix}$

Explanation:

$A^{T}$ , the transpose of , is the result of switching the rows of with the columns.

$A = \begin{bmatrix} 2+ i & 3 - i & i \\ -4 & 1+ i & 3 \end{bmatrix}$ ,

$A^{T} = \begin{bmatrix} 2+ i & -4 \\ 3- i & 1+ i \\i & 3 \end{bmatrix}$

Find the product $AA^{T}$ by multiplying the rows of by the columns of $A^{T}$ ; that is, add the product of the terms in corresponding positions:

$AA^{T}$

$= \begin{bmatrix} 2+ i & 3 - i & i \\ -4 & 1+ i & 3 \end{bmatrix} \begin{bmatrix} 2+ i & -4 \\ 3- i & 1+ i \\i & 3 \end{bmatrix}$

$= \begin{bmatrix}( 2+ i )( 2+ i ) + (3 - i) (3 - i)+ i(i) & ( 2+ i )( -4 ) + (3 - i) (1+i)+ i(3) \\ ( -4 )( 2+ i ) + (1+i)(3 - i)+ 3(i) & -4(-4) +(1+ i)(i+1) + 3(3) \end{bmatrix}$

$= \begin{bmatrix}10-2i & -4+i \\ -4+i & 25+2i \end{bmatrix}$

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Example Question #731 : Linear Algebra

Always, sometimes, or never: AB = BA .

Give the answer for both square and nonsquare matrices.

Possible Answers:

Square: Always

Nonsquare: Sometimes

Square: Sometimes

Nonsquare: Sometimes

Square: Always

Nonsquare: Never

Square: Sometimes

Nonsquare: Never

Square: Never

Nonsquare: Never

Correct answer:

Square: Sometimes

Nonsquare: Never

Explanation:

The statement cannot be true for nonsquare matrices. For to be defined, the number of columns in must be equal to the number of rows in ; for to be defined, the reverse must hold. It follows that and must be $m \times n$ and $n \times m$ matrices, respectively.

The product of two matrices has the same number of rows as the former matrix and the same number of columns as the latter. Therefore, is an $m \times m$ matrix and is an $n \times n$ matrix. If $m \ne n$ , then and do not even have the same dimensions. Therefore, AB = BA is always false for nonsquare matrices.

We now show that AB = BA for some, but not all square matrices. If A = B , it easily follows that AB = BA , since AB = A A = BA . Now let

$A = \begin{bmatrix} 1 & 1 \\ 0& 1 \end{bmatrix}$ and $B =\begin{bmatrix} 1 & 0\\ -1 & 1 \end{bmatrix}$ .

The products are

$AB = \begin{bmatrix} 0 & 1 \\ -1 & 1 \end{bmatrix}$

and

$BA = \begin{bmatrix} 1 & 1 \\ -1 & 0 \end{bmatrix}$

$AB \ne BA$ . Since at least one case exists in which AB = BA and at least one case exists in which $AB \ne BA$ , the statement AB = BA is sometimes true for square matrices.

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Linear Algebra : Linear Algebra

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #42 : Matrices

Example Question #731 : Linear Algebra

Example Question #732 : Linear Algebra

Example Question #733 : Linear Algebra

Example Question #734 : Linear Algebra

Example Question #735 : Linear Algebra

Example Question #51 : Matrix Matrix Product

Example Question #732 : Linear Algebra

Example Question #732 : Linear Algebra

Example Question #731 : Linear Algebra

All Linear Algebra Resources

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